Question

Show that the product of perpendiculars on the line \[\frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta = 1\]  from the points \[( \pm \sqrt{a^2 - b^2}, 0) \text { is }b^2 .\]

Answer 1

Let 

\[d_1 \text { and } d_2\] be the perpendicular distances of line \[\frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta = 1\]  from points \[\left( \sqrt{a^2 - b^2}, 0 \right) \text { and } \left( - \sqrt{a^2 - b^2}, 0 \right)\] ,respectively.

\[\therefore d_1 = \left| \frac{\frac{\sqrt{a^2 - b^2}}{a}cos\theta - 1}{\sqrt{\frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2}}} \right| = b \left| \frac{\sqrt{a^2 - b^2}cos\theta - a}{\sqrt{a^2 \sin^2 \theta + b^2 \cos^2 \theta}} \right|\]

Similarly,

\[d_1 = \left| \frac{- \frac{\sqrt{a^2 - b^2}}{a}cos\theta - 1}{\sqrt{\frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2}}} \right| = b \left| \frac{- \sqrt{a^2 - b^2}cos\theta - a}{\sqrt{a^2 \sin^2 \theta + b^2 \cos^2 \theta}} \right| = b \left| \frac{\sqrt{a^2 - b^2}cos\theta + a}{\sqrt{a^2 \sin^2 \theta + b^2 \cos^2 \theta}} \right|\]

Now,

\[d_1 d_2 = b \left| \frac{\sqrt{a^2 - b^2}cos\theta - a}{\sqrt{a^2 \sin^2 \theta + b^2 \cos^2 \theta}} \right| \times b \left| \frac{\sqrt{a^2 - b^2}cos\theta + a}{\sqrt{a^2 \sin^2 \theta + b^2 \cos^2 \theta}} \right|\]

\[ \Rightarrow d_1 d_2 = b^2 \left| \frac{\left( a^2 - b^2 \right) \cos^2 \theta - a^2}{a^2 \sin^2 \theta + b^2 \cos^2 \theta} \right|\]

\[ \Rightarrow d_1 d_2 = b^2 \left| \frac{a^2 \left( \cos^2 \theta - 1 \right) {- b}^2 \cos^2 \theta}{a^2 \sin^2 \theta + b^2 \cos^2 \theta} \right|\]

\[ \Rightarrow d_1 d_2 = b^2 \left| \frac{- a^2 \sin^2 \theta {- b}^2 \cos^2 \theta}{a^2 \sin^2 \theta + b^2 \cos^2 \theta} \right| = b^2 \left| \frac{a^2 \sin^2 \theta {+ b}^2 \cos^2 \theta}{a^2 \sin^2 \theta + b^2 \cos^2 \theta} \right| = b^2\].